PUSH AND RELEASE OF A PISTON
Statement
Inside a vertical cylinder of 0.01 m2 cross-section there is 0.01 kg of nitrogen closed by a 5 kg steel piston
at the top. Consider the following process: 1) by a suitable force, the piston is gently moved so as to
reduce the initial volume of the gas by 10%; 2) the applied force is suddenly removed. One wants to:
a) Sketch of the evolution in a z-t (height-time) diagram.
b) Determine the thickness of the piston and its initial height.
c) State of the system at each point and a p-V (pressure-volume) process diagram.
d) Energy change between states, and heat and work transferred through the frontier.
e) T-s (temperature-entropy) process diagram.
f) Entropy change and entropy generation for every system between consecutive states, and along the
whole process.
Dentro de un cilindro vertical de 0,01 m2 de sección hay 0,01 kg de nitrógeno encerrado con un émbolo
superior de 5 kg de acero. Considérese la siguiente evolución: 1) mediante las fuerzas apropiadas se
obliga al pistón a reducir lentamente en un 10% el volumen ocupado por el gas, y 2) se libera el anclaje y
se permite el libre movimiento del émbolo. Se pide:
a)
Esquema de la evolución en un diagrama altura-tiempo.
b)
Determinar el espesor del émbolo y su altura inicial.
c)
Valores p-V-T en los estados de equilibrio considerados.
d)
Variación de energía entre los estados antedichos y para todo el ciclo, así como calor y trabajo
e)
f)
transferidos entre los sistemas involucrados.
Diagrama T-s de la evolución.
Variación de entropía y generación de entropía para todos los sistemas entre los estados antedichos
y para todo el ciclo.
Solution
a)
Sketch of the evolution in a z-t (height-time) diagram.
Most problems are not as simple as the previous exercises, and the solution is not immediate but requires
some development, mixing appropriately comments on notation and assumptions, analytical deductions
(formulation) and numerical computations. To better grasp the problem by the person solving it, and to
better transmit the information to the persons using it, it is important to start by a sketch of the system and
its evolution, introducing also the nomenclature used, as here in Fig. 1.
Fig. 1. Time evolution of the piston: 1) initial state, 2) 10% compression, 3) maximum
expansion, 4) mechanical equilibrium, 5) thermal equilibrium.
Push and release of a piston
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First the initial conditions are studied. From the problem statement we should understand that the piston
is initially in mechanical equilibrium so that the pressure in the gas is p1=p0+mPg/A, where p0 is the
ambient atmospheric pressure (assumed as 100 kPa as reasoned in the Introduction), mP is the mass of the
piston, g the gravitational acceleration (assumed as 9.8 m/s2 as reasoned in the Introduction) and A the
cross-section area A=D2/4. In absence of more details, we also take T0=288 K (15 ºC) as reasoned in the
Introduction.
b)
Determine the thickness of the piston and its initial height.
The density of steel is assumed available: =7900 kg/m3 (from Thermal data), so the thickness is
LP=mP/(PA)=0.064 m.
We obtained the height of the piston from the ideal gas equation of state pAz=mRT with with T1=288 K
and p1=p0+mPg/A=105 kPa, what yields z1=mRT/(p1A)=0.815 m.
c)
State of the system at each point and a p-V (pressure-volume) process diagram.
From 1 to 2 a slow compression takes place, thence enough time for heat transfer may be assumed and the
process approximated as isothermal, so that pV=constant, and with z2=(V2/V1)z1, z2=0.9∙0.815=0.734 m,
yields p2=p1(V1/V2)=117 kPa with T2=288 K.
The evolution after release of the piston in state 2 is rather complicated, but the final state of
thermodynamic equilibrium (state 5 in Fig. 1) is trivial because temperature and pressure are recovered
from the initial state, so that p5=105 kPa, z5=0.815 m, with T5=288 K.
Point 3, the maximum expansion, is obtained assuming that there is no time for heat transfer, and that the
friction force Ff between piston and cylinder is small. Upon integration of the momentum equation for the
piston:
z z z
F  mP 
z
 Fdx  mP 
zdz  mP 
zzdt  12 mP z 2 
zd
i
pA  p0 A  mP g  Ff dz
(1)
from state 2 to 3, both with zero speed, neglecting Ff, and with pz=constant, one gets:
z3
0 
z2

1
 z12 
 z2 
  z3
p2   Adz  ( p0 A  mP g )( z3  z2 )  p2 z2 
  ( p0 A  mP g )( z3  z2 ) (2)
 z 
 1  
This is one equation with one unknown z3 that must be numerically or graphically evaluated since it is not
explicitly solved algebraically because of the power . To do it by hand, once the known values
substituted:
0  2910 
1891
 1049z3
z30.4
(3)
it is better to start by the expected solution; if it were a linear oscillation around z1=0.815 m from
z2=0.734 m, it would reach z5=0.815+(0.8150.734)=0.896 m, so this is a good start, but we get for the
right-hand-side of (3) a value of 6 (the unit is J), instead of 0; trying now with a close value e.g. 0.85 we
Push and release of a piston
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get 0.2 instead of 0, and interpolating linearly, the zero would correspond to z 3=0.851 m. More accurately
result is z3=0.852 m; from pV=constant one gets p3=95 kPa, and from the equation of state T3=271 K.
Just to help visualise the problem, Fig. 2 presents the function to be cancelled:
f(z)=29101891/z0.41049z.
Fig. 2. Plot of the function to be cancelled in (3), i.e. f(z)= 29101891/z0.41049z. The
trivial solution is z=z2=0.734 m, as in (2), and the one looked for is z=z3=0.852 m.
Point 4 corresponds to the mechanical equilibrium, which is quickly reached even for small-friction
systems, far quicker than the thermal equilibrium when gases are participating, that takes a very long
time. Because of mechanical equilibrium of the piston, p4=p0+mPg/A=105 kPa, and, assuming that heat
transfer is still under-developed, pV=constant, because there is not friction inside the system (gas);
thence z4=z2(p2/p4)= 0.734(117/105)1.4=0.791 m and T4=279 K.
Point 5 coincides with the initial state, as said before. The p-V-T values (really z-T-p values) are
summarised in Table 1, and the evolution in the p-V diagram shown in Fig. 3.
Table 1. Values of state variables at key points in the evolution of the gas.
state
z [m] T [K] p [kPa]
1
0.815
288
105
2
0.734
288
117
3
0.852
271
95
4
0.791
279
105
5
0.815
288
105
Fig. 3. Evolution in the p-V diagram. From 1 to 2 isothermal, from 2 to 3 adiabatic
oscillations dumped until mechanical equilibrium at point 4, from 4 to 5 isobaric
tempering.
d) Energy change between states, and heat and work transferred through the frontier.
The energy change between two states for the trapped gas is U=mcvT where the thermal capacity at
constant volume for the gas, cv, is assumed to be known, usually through the Mayer relation cp-cv=R and
Push and release of a piston
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tabulated cp values (see Thermal data): cv=cp-R=1040-297=743 J/(kg∙K). The work received by the system
(the trapped gas) is computed by W=∫pdV, once the function p=p(V) is determined by the actual process.
The heat received by the system is computed with Q=EW. The results obtained are summarised in
Table 2 and analysed below.
Table 2. Energy change, and work and heat received in the gas system.
process E [J] W [J] Q [J]
1-2
0
90.1 -90.1
2-4
-63.4 -63.4
0
4-5
63.4
-25.3 88.7
1-5
0
1.4
-1.4
Obviously E=W+Q. Also, because of the cyclic process, E15=0, but W150 and consequently Q150.
Since the work done is the area between the p=p(V) path and the abscissas, the work in the cycle is the
area enclosed by the closed path (the little triangle-wise 1-2-4-5 in Fig. 3), that is positive (received by the
system) if the cycle is run anti-clockwise, as in this exercise, and negative in the contrary. It will be
shown in Chap. 3 that this cycle represents a refrigeration machine, capable of extracting heat (process 4
to 5) from a system below ambient temperature, at the expense of some external work. It is also
interesting to note that in the process 1-2, from the 90.1 J received by the gas, only 4.6 J were actually
consumed by the external force pushing the piston; the rest were supplied by the decreasing potential
energy of the piston, mEgz=4.0 J, and primarily by the ambient atmosphere, p0V=81.5 J. From this 4.6
J directly applied, 1.4 J are used for the refrigeration effect (Table 2) and the other 3.2 J dissipated in the
friction between piston and cylinder.
e) T-s (temperature-entropy) process diagram.
The generic entropy change between two states for the trapped gas is S2-S1=m(cpln(T2/T1)-Rln(p2/p1)). The
numerical values are computed below, but the sketch is easy: an isothermal entropy decrease (heat
flowing out), isentropic oscillations, and a final isobaric tempering to re-gain initial conditions.
Fig. 4. Evolution in the T-S diagram. Note that the origin of S is arbitrary (here, S0=0.5 J/K is taken, to
centre the process path).
Push and release of a piston
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f) Entropy change and entropy generation for every system between consecutive states, and along the
whole process.
The generic entropy change between two states for the system, S, has been quoted above, and for the
atmosphere it is Senv=Q0/T0 (with Sgen,env=0 because it is assumed to be a reversible heat source). But
there is a singularity with this modelling; as explained in Exercise 2: we must add a third thermodynamic
system, the frontier, to account for all physical processes, since, with the uniform-temperature model for
the system and the environment, entropy is generated at the frontier. To skip that singularity, we follow
here the trick of extending the environment a little inside the system, to include that dissipation; thence
Senv=Q0/T with Sgen,env>0. The generation of entropy in the universe is now Sgen=S+Senv. The results
are presented in Table 3.
From 1 to 2, S12=m(cpln(T2/T1)-Rln(p2/p1))=0.01(1040ln(288/288)-297ln(117/105))=0.313 J/K. For
the environment receives Senv=Q0/T0=90.1/288=0.313 J/K.
From 2 to 4, gas evolution has been assumed isentropic, S=0, but the entropy of the environment has
increased due to friction dissipation. Although there is no time from 2 to 4 for the heated pieces in friction
(piston and cylinder walls) to evacuate the energy to the atmosphere (and not to the gas inside), let us
assume the model is appropriate, and compute Senv= Q0/T0=W24/T0=63.4/288=0.222 J/K.
From 4 to 5, S45=m(cpln(T5/T4)-Rln(p5/p4))=0.01(1040ln(288/279)-297ln(105/105))= 0.313 J/K. For the
environment receives Senv=Q0/T0=88.7/288=0.308 J/K.
Table 3. Entropy change for the system (trapped gas) and for the environment (including the
frontier), and global entropy generation.
Gas
Environment
Universe
process
S
∫dQ/T Sgen
S
∫dQ/T Sgen
S ∫dQ/T Sgen
[J/K] [J/K] [J/K] [J/K] [J/K] [J/K] [J/K] [J/K] [J/K]
1-2
-0.313 -0.313
0
0.313 0.313
0
0
0
0
2-3-4
0
0
0
0.222
0
0.222 0.222
0
0.222
4-5
0.313 0.313
0
-0.308 -0.313 0.005 0.005
0
0.005
1-2-3-4-5
0
0
0
0.227
0
0.227 0.227
0
0.227
Notice that entropy generation is always positive, both at a single step and for the whole process, whereas
entropy changes of a non-isolated system may be positive or negative.
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